What is the integral of x(x-1)^{99} ?

The integral of x(x-1)^{99} is ((x-1)^{101})/(101)+((x-1)^{100})/(100)+C

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integral of x(x-1)^{99}

\int x (x - 1)^{99} d x

\frac{( x - 1 ) ^{101}}{101} + \frac{( x - 1 ) ^{100}}{100} + C

Solution steps

\int x (x - 1)^{99} d x

Apply u-substitution

: \int (u + 1) u^{99} d u

= \int (u + 1) u^{99} d u

Expand

(u + 1) u^{99} : u^{100} + u^{99}

= \int u^{100} + u^{99} d u

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int u^{100} d u + \int u^{99} d u

\int u^{100} d u = \frac{u ^{101}}{101}

\int u^{99} d u = \frac{u ^{100}}{100}

= \frac{u ^{101}}{101} + \frac{u ^{100}}{100}

Substitute back

u = x - 1

= \frac{( x - 1 ) ^{101}}{101} + \frac{( x - 1 ) ^{100}}{100}

Add a constant to the solution

= \frac{( x - 1 ) ^{101}}{101} + \frac{( x - 1 ) ^{100}}{100} + C

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What is the integral of x(x-1)^{99} ?
The integral of x(x-1)^{99} is ((x-1)^{101})/(101)+((x-1)^{100})/(100)+C